CHAPTER 6
1.
Assume a yield to maturity of 8 percent. Compute the duration
for the following bonds. Assume $100 par values. For the 12%
coupon bond, compute the duration using the two duration
formulas. Which formula is easiest to compute?
(a) 10 years, zero coupon
(b) 10 years, 8 percent coupon
(c) 10 years, 12 percent coupon
YTM = 8% ; par = $100 ; DUR = ?
a. N = 10 ; C = 0 ; DUR = 10
Duration equals the maturity of 10.
b. N = 10; C = 8%
DUR par bond =
DUR =
(1 + y)[1  (1 + y )-n ]
y
(1.08)[1  (1.08 )-10]
= 7.2469
0.08
c. N = 10; C = 12%
 (1  (1  y )-n )   Par 
P= c 
+
n
y

  (1 + y ) 
 (1  (1.08 )-10   100 
P = 12 
= 126.84
+
10 
0.08

  (1.08 ) 
 par  c/y 
 c(1 + y)  1  (1 + y )-n 
 + n

 y 
y
(1 + y )n 




(1) DUR =
P
-10
100  12/ 0.08 
12(1.08)  1  (1.08 ) 
 0.08   0.08  + 10 (1.08 )10 




DUR =
126.84
DUR  6.7442
(1.08)[1  (1.08 )-10]
(2) DUR par bond =
= 7.2469
0.08
 c/P 
DUR non par = n  [n  DUR par] 
 y 
12 / 126.84 
DUR non par = 10  10  7.24689
 0.08 
DUR non par  6.7442
2.
In problem 1, assume that yields change from 8 to 9 percent.
Work out the exact change in price and compare it with the
change in price predicted by duration. Explain the difference.
Assume $100 par values.
a.
(P1  P0)
=
P0
(42.24  46.32)
= - 0.0880
46.32
- (Change in yield)(DUR) = (-0.01)(10) = -0.10
The duration approximation is slightly more than 1 percent
too large in absolute value.
b.
(P1  P0)
=
P0
(93.58  100)
= - 0.0642
100
- (Change in yield)(DUR) = (-0.01)(7.2468) = -0.0725
The duration approximation is 0.83 percent too large in
absolute value.
c.
( P1  P0 )
P0
=
(119.25  126.84)
= - 0.0598
126.84
- (Change in yield)(DUR) = (-0.01)(6.7442) = -0.0674
The duration approximation is 0.76 percent too large in
absolute value.
3.
Compute the duration of a portfolio composed of a ten-year,
zero coupon bond and a ten-year, 8 percent coupon bond. Yield
to maturity is 8%. For simplicity assume each bond has a par
value of $100. Suppose that equal dollar amounts are invested
in the two bonds.
Bond 1:
N = 10; C = 0; FV = 100; I/YR = 8; PV = ?
PV = 46.319
DUR zero coupon bond = n = 10
Bond 2:
N = 10; C = 8; FV = 100; I/YR = 8; PV = ?
PV = 100
(P1)(DUR) + (P2)( DUR 2)
P1 + P2
( 46.319 )(10) + (100)( 7.247 )
= 8.1185
DUR portfolio=
146.319
DUR portfolio =
4.
A perpetual bond has a coupon of $6 and a yield to maturity of
6 percent. Work out the actual percentage change in price and
the duration approximation in the following three cases:
(a) The yield decreases by 1 percent
(b) The yield increases by 1 percent
(c) The yield increases by 8 percent.
P0 =
6
= $100.00
0.06
DUR perpetual =
a. P1 =
1 + y 1.06
=
= 17.67
y
0.06
6
= $120.00
0.05
% change =
120  100
= 0.2 = 20%
100
%P  ( Duration )y  17.67  0.01  17.67%
b. P1 =
6
= $85.71
0.07
% change =
85.71  100
= - 0.1429 = - 14.29%
100
%P  ( Duration )y  17.67  (0.01)  17.67%
6
= $42.86
c. P1 =
0.14
% change =
42.86  100
= - 0.5714 = - 57.14%
100
%P  ( Duration )y  17.67  (0.08)  141.36%
8.
Suppose that the interest rate on all bonds is 6%. Which is a
correct statement?
(a) The duration of a par bond can never be greater than 18 for
any maturity.
(b) The duration of a premium bond can never be greater than
17, even for long maturities.
(c) The duration of a 15-year Treasury strip exceeds the
duration of all par bonds.
(d) The duration of a discount bond can never exceed 17.
(e) None of the above.
DUR
c=0
17.67 =
1 y
y
par
Premium
1
1
(a) is correct .
17.67
M
9.
Suppose all interest rates are 7%. We observe a bond
with a duration of 18. What can we say about the coupon rate
of this bond?
DUR
15.2857 =
1 y
y
1
1
15.2857
This must be a discount bond. Par bonds have a coupon rate of
7%. So, discount bonds have a coupon rate below 7%.
10.
Compute the duration for the following bond. Coupon = $9,
par = $100, maturity = 12 years. The yield to maturity is 10%.
c / P
DUR  N  N  DUR par 

 y 
P = 93.1863
 9 / 93.1863 
DUR  12  12  7.495061

 0.10 
 9 / 93.1863 
 12  4.504939

 0.10 
 12  4.5049390.965807
 12  4.350012
 7.64938.
N